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Question Number 22481 by ajfour last updated on 19/Oct/17

Commented by ajfour last updated on 19/Oct/17

$Q . 22479 (S o l u t i o n)$
	Answered by ajfour last updated on 19/Oct/17

	$W h e n t h e l a d d e r i s a b o u t t o s l i p :$ $I f μ_{1} \neq 0, μ_{2} \neq 0$ $f_{2} = μ_{2} N_{2} = N_{1} . . . . (i)$ $f_{1} + N_{2} = μ_{1} N_{1} + N_{2} = m g$ $\Rightarrow (μ_{1} μ_{2} + 1) N_{2} = m g$ $N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$ $c o n s i d e r i n g τ o r q u e a b o u t c e n t r e :$ $N_{2} (\frac{l}{2}) \cos θ = f_{2} (\frac{l}{2}) \sin θ +$ $f_{1} (\frac{l}{2}) \cos θ + N_{1} (\frac{l}{2}) \sin θ$ $\Rightarrow N_{2} (1 - μ_{2} \tan θ) = N_{1} (μ_{1} + \tan θ)$ $. . . . . (i i i)$ $\Rightarrow I f μ_{1} = 0, μ_{2} \neq 0 t h e n$ $f r o m (i i) w e g e t N_{2} = m g, a n d$ $N_{1} = μ_{2} m g . T h e n f r o m (i i i) w e g e t$ $m g (1 - μ_{2} \tan θ) = N_{1} \tan θ$ $\Rightarrow m g - N_{1} \tan θ = N_{1} \tan θ$ $o r N_{1} \tan θ = \frac{m g}{2} .$ $I f μ_{1} \neq 0, μ_{2} = 0$ $f o r t h e a b o u t t o s l i p c a s e$ $f_{2} = 0 \Rightarrow N_{1} = 0, θ = \frac{π}{2}$ $N_{2} = m g .$ $H e n c e (3), (4) a r e c o r r e c t .$
	Commented by Tinkutara last updated on 19/Oct/17

	$Thank you very much Sir!$
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